Expert: Paul Klarreich - 3/8/2008

Question
A function f is defined by

f(x)= 1/3 + (2/3^2)x + (3/3^3)x^2 + ...+ [(n+1)/3^(n+1)]x^n
+....
for all x in the interval of convergence of the given power series.

a, Find the interval of convergence for this power series. Show he work that leads to your answer.

b, Find lim (f(x)- 1/3) / x as x --> 0

for a, i got -3 < x <3
and for b, i got - infinity but i don't think that's right.

can you please help me with part b?

i really appreciate your help!
Thank you.


Answer
Questioner:   Anna
Category:  Calculus
Private:  No
 
Subject:  cal bc
Question:  A function f is defined by

f(x)= 1/3 + (2/3^2)x + (3/3^3)x^2 + ...+ [(n+1)/3^(n+1)]x^n +....
for all x in the interval of convergence of the given power series.

a, Find the interval of convergence for this power series. Show he work that leads to your answer.

b, Find lim (f(x)- 1/3) / x as x --> 0

for a, i got -3 < x <3
and for b, i got - infinity but i don't think that's right.

can you please help me with part b?

i really appreciate your help!
Thank you.
.....................................
Hi, Anna,

If  f(x) = SUM [(n+1)/3^(n+1)]x^n, then use the ratio test on consecutive terms:

|a[n+1]|   | [(n+2)/3^(n+2)]x^(n+1) |
-------- = --------------------------- =
|a[n]|     | [(n+1)/3^(n+1)]x^n |

a[n+1]   |[(n+2)/3]x |
------ = -------------
a[n]     | [(n+1)  |

Now you want lim(n->inf) of that:
 n + 2  |x|       |x|
= ------ ---  -->  ---
 n + 1   3         3

And you need that to be < 1, so  | x | < 3, or -3 < x < 3

Yes, you did OK on that part.
..................

b, Find lim (f(x)- 1/3) / x as x --> 0

Looks to me as if that is just


1/3 + (2/3^2)x + (3/3^3)x^2 + ...+ [(n+1)/3^(n+1)]x^n +.... minus 1/3, which is:

(2/3^2)x + (3/3^3)x^2 + ...+ [(n+1)/3^(n+1)]x^n +....

Now dividing by x, that is:

(2/3^2) + (3/3^3)x + ...+ [(n+1)/3^(n+1)]x^(n-1) +....

and as x --> 0, all the terms drop out except the first, so the answer is

2/3^2